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    Hi I'm wondering if somebody would be able to help me with this question, cant get my head around it...

    Two particles of masses 6m and m are attached to the ends A and B of a light inextensible string which passes over a smooth pulley C at the top of a fixed rough plane inclined at 300 to the horizontal. The particles are at rest with the 6m mass in contact with the plane and the mass m hanging freely so that AC is along the greatest slope of the plane and BC is vertical.

    1) Find the frictional force between the 6m and the plane (3 marks).
    2) Find the range of values for the coefficient of friction for equilibrium to be possible. (4 Marks)

    Many Thanks
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    (Original post by SuperHuman10)
    Hi I'm wondering if somebody would be able to help me with this question, cant get my head around it...

    Two particles of masses 6m and m are attached to the ends A and B of a light inextensible string which passes over a smooth pulley C at the top of a fixed rough plane inclined at 300 to the horizontal. The particles are at rest with the 6m mass in contact with the plane and the mass m hanging freely so that AC is along the greatest slope of the plane and BC is vertical.

    1) Find the frictional force between the 6m and the plane (3 marks).
    2) Find the range of values for the coefficient of friction for equilibrium to be possible. (4 Marks)

    Many Thanks
    start by drawing a diagram


    http://www.thestudentroom.co.uk/showthread.php?t=403989
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    Name:  ImageUploadedByStudent Room1421953585.491180.jpg
Views: 75
Size:  111.3 KB

    That is my diagram. Do I need to form sim equations or resolve vertically?


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    (Original post by SuperHuman10)
    Name:  ImageUploadedByStudent Room1421953585.491180.jpg
Views: 75
Size:  111.3 KB

    That is my diagram. Do I need to form sim equations or resolve vertically?


    Posted from TSR Mobile
    good start

    firstly when the particles move, equilibrium is broken
    can you find an expression for the friction by resolving perpendicular to the plane?
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    R = 6mg x cos(30)


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    (Original post by SuperHuman10)
    R = 6mg x cos(30)


    Posted from TSR Mobile
    correct

    so you now have an expression for the friction.

    next 2 separate equations of motion for each object


    PS:If you are replying please use the reply with quote button otherwise I cannot see you unless I refresh the screen
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    (Original post by TeeEm)
    correct

    so you now have an expression for the friction.

    next 2 separate equations of motion for each object


    PS:If you are replying please use the reply with quote button otherwise I cannot see you unless I refresh the screen
    But there are two unknowns? The coefficient of friction and friction?


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    (Original post by SuperHuman10)
    But there are two unknowns? The coefficient of friction and friction?


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    no coefficient of friction to start with

    T= friction
    T = mg
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    (Original post by TeeEm)
    no coefficient of friction to start with

    T= friction
    T = mg
    What about 6mg sin(30)?


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    (Original post by SuperHuman10)
    What about 6mg sin(30)?


    Posted from TSR Mobile
    you are correct

    I do not have the picture in front of me


    T=mg from the hanging object

    T = friction +6mgsin30

    then you find friction (as a total)
 
 
 
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